Cauchy principal value and displacing singularity

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Let $x\in\mathbb{R}$, let $\alpha\in\mathbb{R}_{>0}$, and consider the integral \begin{equation} \int_{-\infty}^\infty\mathrm{d}y\,\frac{\exp(ixy)}{y^2-\alpha}. \end{equation} The integral as such diverges because of the simple poles on the real axis. In the actual phrasing of the question (this is homework) the teacher writes $\alpha^+:=\lim_{\epsilon\downarrow0}\alpha+i\delta$ instead of $\alpha$, seemingly suggesting that we should actually calculate the integral \begin{equation} \int_{-\infty}^\infty\mathrm{d}y\,\frac{\exp(ixy)}{y^2-(\alpha+i\delta)}, \end{equation} where $\delta>0$, and then take the limit $\delta\downarrow0$. However that is not how the question is stated, and I remembered that even though the first integral is divergent, the Cauchy principal value of the integral \begin{equation} \lim_{\delta\downarrow0}\left(\int_{-\infty}^{-\sqrt{\alpha}-\delta}\mathrm{d}y\,\frac{e^{ixy}}{y^2-\alpha}+\int_{-\sqrt{\alpha}+\delta}^{\sqrt{\alpha}-\delta}\mathrm{d}y\,\frac{e^{ixy}}{y^2-\alpha}+\int_{\sqrt{\alpha}+\delta}^\infty\mathrm{d}y\,\frac{e^{ixy}}{y^2-\alpha}\right) \end{equation} is well-defined, so I wanted to calculate this integral. Perhaps slightly naively I was expecting to get the same answer, and the comments around equation (7.39) in this document convinced me, but, when I do the actual calculation, I do not get the same answer. Of course my calculation can contain one or more errors but I feel like it is in order. Is it true that these ways of calculating are equivalent, and did I make a mistake, or are they actually different, and if so, how come? Any help is appreciated.

I should note that this is homework for a physics course.

EDIT To comment on user1952009's comments:

The actual integral we are calculating is \begin{equation} \int_{-\infty}^\infty\mathrm{d}q\,\frac{e^{\frac{ixq}{\hbar}}}{E^+-\frac{q^2}{m}}, \end{equation} where $E^+:=\lim_{\epsilon\downarrow0}$. We have to show that it is equal to \begin{equation} -i\pi\sqrt{\frac{m}{E}}\exp\left(\frac{i|x|\sqrt{mE}}{\hbar}\right), \end{equation} so no, what you said is not what my teacher meant. What she means is calculating the integral \begin{equation} \int_{-\infty}^\infty\mathrm{d}q\,\frac{e^{\frac{ixq}{\hbar}}}{E+i\epsilon-\frac{q^2}{m}}, \end{equation} and take the limit $\epsilon\downarrow0$, because this is the only wy to get the desired answer, and because one of the TA's the limit is there to shift the poles (even though that is not true because the limit just exists and is equal to $E$, so really one cannot show this is, because it is simply not true). Your comment is useful however because the PV I calculated using the definition I mentioned precisely gives me what would come out of the two integrals with the singularities shifted, as you said one can prove. What do you think? And how does one go about proving what you said?

Answer 1

The Cauchy Principal Value is given by

$$\begin{align} \text{PV}\left(\int_{-\infty}^\infty \frac{\exp(ixy)}{y^2-\alpha}\,dy\right)&=\lim_{\delta\to 0^+}\left(\int_{-\infty}^{-\sqrt\alpha -\delta} \frac{e^{ixy}}{y^2-\alpha}\,dy+\int_{-\sqrt\alpha +\delta}^{\sqrt\alpha-\delta} \frac{e^{ixy}}{y^2-\alpha}\,dy+\int_{\sqrt\alpha +\delta}^\infty \frac{e^{ixy}}{y^2-\alpha}\right) \end{align}$$

Using Cauchy's Integral Theorem, for $x>0$ this integral is given by

$$\begin{align} \text{PV}\left(\int_{-\infty}^\infty \frac{\exp(ixy)}{y^2-\alpha}\,dy\right)&=\lim_{\delta\to0^+}\int_0^\pi \frac{e^{ix(-\sqrt\alpha +\delta e^{i\phi})}\,i\delta e^{i\phi}}{\delta e^{i\phi}(-2\sqrt\alpha +\delta e^{i\phi})}\,d\phi+\lim_{\delta\to0^+}\int_0^\pi \frac{e^{ix(-\sqrt\alpha +\delta e^{i\phi})}\,i\delta e^{i\phi}}{\delta e^{i\phi}(2\sqrt\alpha +\delta e^{i\phi})}\,d\phi\\\\ &=-\pi \frac{\sin(\sqrt \alpha x)}{\sqrt\alpha} \end{align}$$

where neither pole is enclosed by the applicable contour.

Similarly for $x<0$, we have

$$\text{PV}\left(\int_{-\infty}^\infty \frac{\exp(ixy)}{y^2-\alpha}\,dy\right)=\pi \frac{\sin(\sqrt \alpha x)}{\sqrt\alpha} $$

Note that this Cauchy Principal Value integral differs conceptually from the function $F(x,\alpha)$ as given by

$$\begin{align} F(x,\alpha)&=\lim_{\epsilon\to 0^+}\int_{-\infty}^\infty \frac{e^{ixy}}{y^2-(\alpha +i\epsilon)}\tag 1 \end{align}$$

In $(1)$, the poles are in the first and third quadrants. Therefore, only one, but not both of the poles is enclosed in a contour in the upper-half plane or the lower half-plane. Then, applying the Residue Theorem in the case for which $x>0$, we obtain

$$F(x,\alpha)=i\pi \frac{e^{i\sqrt\alpha x}}{\sqrt \alpha}$$

while in the case for which $x<0$, we obtain

$$F(x,\alpha)=-i\pi \frac{e^{-i\sqrt\alpha x}}{\sqrt \alpha}$$